Question: Simplify and expand the following expression: $ \dfrac{1}{4t - 8}+ \dfrac{3}{2t - 18}+ \dfrac{3}{t^2 - 11t + 18} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{1}{4t - 8} = \dfrac{1}{4(t - 2)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{3}{2t - 18} = \dfrac{3}{2(t - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{t^2 - 11t + 18} = \dfrac{3}{(t - 2)(t - 9)}$ Now we have: $ \dfrac{1}{4(t - 2)}+ \dfrac{3}{2(t - 9)}+ \dfrac{3}{(t - 2)(t - 9)} $ The least common multiple of the denominators is: $ 8(t - 2)(t - 9)$ In order to get the first term over $8(t - 2)(t - 9)$ , multiply by $\dfrac{2(t - 9)}{2(t - 9)}$ $ \dfrac{1}{4(t - 2)} \times \dfrac{2(t - 9)}{2(t - 9)} = \dfrac{2(t - 9)}{8(t - 2)(t - 9)} $ In order to get the second term over $8(t - 2)(t - 9)$ , multiply by $\dfrac{4(t - 2)}{4(t - 2)}$ $ \dfrac{3}{2(t - 9)} \times \dfrac{4(t - 2)}{4(t - 2)} = \dfrac{12(t - 2)}{8(t - 2)(t - 9)} $ In order to get the third term over $8(t - 2)(t - 9)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{3}{(t - 2)(t - 9)} \times \dfrac{8}{8} = \dfrac{24}{8(t - 2)(t - 9)} $ Now we have: $ \dfrac{2(t - 9)}{8(t - 2)(t - 9)} + \dfrac{12(t - 2)}{8(t - 2)(t - 9)} + \dfrac{24}{8(t - 2)(t - 9)} $ $ = \dfrac{ 2(t - 9) + 12(t - 2) + 24} {8(t - 2)(t - 9)} $ Expand: $ = \dfrac{2t - 18 + 12t - 24 + 24}{8t^2 - 88t + 144} $ $ = \dfrac{14t - 18}{8t^2 - 88t + 144}$ Simplify: $ = \dfrac{7t - 9}{4t^2 - 44t + 72}$